Pre Calculus Understanding Exponents

At this point, we are quite used to using exponents; we have seen them a bunch, and we just did them a whole lot when we were working with polynomials. We know that an exponent is just a shorthand way to express repeated multiplication. This lesson will show us how to work with any kind of exponent - any real number. This is the lesson that we are actually going to see how we build the rules of exponents, where they come from, why we can trust in them, and also how we can make them ourselves, if we ever forget the rules.

Hi, When should we use the "absolute value" sign (for questions that ask us to change the expression from the root form to the exponential form)?

Thank you very much:)

1 answer

Last reply by: Professor Selhorst-JonesTue Aug 4, 2015 8:49 AM

Post by Duy Nguyenon August 4, 2015

Hi,suddenly I have unsupported format issue with educator.comThe video will not play. I am using a Mac. Please advise.Thank you very much

Understanding Exponents

At heart, exponentiation is repeated multiplication. By definition, for any number x and any positive integer a,

xa =

x·x ·x …x ·x

.

We can expand on this fundamental idea to see how exponentiation can work with numbers that aren't just positive integers.

Through multiplication, we can combine numbers that have the same exponent base:

xa ·xb = xa+b.

We can consider exponentiation acting upon exponentiation:

(xa)b = xa·b.

We can look at raising two numbers to the same exponent:

xa ·ya = (xy)a.

For any number at all, raising it to the 0 turns it into 1:

x0 = 1.

Raising a number to a negative will "flip" it to its reciprocal:

x−a =

1

xa

,

⎛⎝

x

y

⎞⎠

=

⎛⎝

y

x

⎞⎠

.

Because of the above, we see that a denominator is effectively a negative exponent. This means if we have a fraction where the numerator and denominator have the same base, we can subtract the denominator's exponent from the numerator's exponent:

xa

xb

= xa−b.

Raising a number to a fraction is the equivalent of taking a root:

x[1/2] = √x, x[1/3] =

3

√

x

, x[1/n] =

n

√

x

.

If we want to find the value of raising something to an irrational number, we can find a decimal approximation of the true value by just using many decimals from our irrational number:

8π = 83.1415926… ≈ 83.14159.

The more accurate we need our approximation to be, the more decimals we can use from the irrational number.

Understanding Exponents

Simplify each of the below expressions.

x4·x9 t ·t3 ·t7 ·t8 ·t−5

The important rule here is one of the most fundamental rules of exponents: xa ·xb = xa+b. [This rule is true for any variable or expression, we just use x for convenience.]

For the first expression, x4 ·x9, the rule says that because they are directly multiplying each other and have the same base, we add the exponents.

x4 ·x9 = x4+9 = x13

Things work similarly for the second expression, we just need to repeat the process. We could add the first two together, then the next to the result, then the next, and so on. Alternatively, and more easily, since we're just going to wind up adding all the exponents together, we can add them all together in a single step:

t ·t3 ·t7 ·t8 ·t−5 = t1 + 3 + 7 + 8 −5 = t14

[If you're not sure where the 1 in the sum comes from, it's because a t on its own can be seen as t1.]

x13 t14

Simplify each of the below expressions.

(x3)2 ( ( ( t2 )5 )[1/4] )2

The important rule here is that if you exponentiate something that's already been exponentiated, the exponents multiply: (xa)b = xa·b. [This rule is true for any variable or expression, we just use x for convenience.]

For the first expression, (x3)2, the rule says that because we're raising it to the 2 after already raising it to the 3, we multiply the numbers:

(x3)2 = x3 ·2 = x6

Things work similarly for the second expression, we just need to repeat the process. We could start with the in-most part of the expression, multiply those exponents, then do that again with the next exponent, then again with the next. Alternatively, and more easily, since we're just going to wind up multiplying them all together, we can multiply them all together in a single step:

( ( ( t2 )5 )[1/4] )2 = t2 ·5 ·[1/4] ·2 = t5

x6 t5

Simplify each of the below expressions.

x0 47190 ( a + b10 + 28)0

The important rule here is that anything raised to the 0 becomes 1: x0 = 1. [This rule is true for any variable or expression, we just use x for convenience.]

This is true for anything: if a number or a variable or an expression is raised to the 0, it becomes 1: 47190 = 1.

This is true for the last expression as well: ( a + b10 + 28)0 = 1. It might be a little bit confusing because there are multiple things there, but think about it like this: no matter what (a + b10 + 28) comes out to be, it comes out to be a number. When you raise any number to the 0, it becomes 1. So, because the expression is in parentheses, we're raising the whole expression to the 0, so we get 1.

1 1 1

Simplify each of the below expressions.

x−4

⎛⎝

4

7

⎞⎠

−2

⎛⎝

u−4

v−6w2

⎞⎠

−3

The important rule here is that a negative exponent causes a fraction to "flip". If something isn't already written as a fraction, remember, you can always put it over 1, then flip the top and bottom. From the rules in the lesson, we have x−a = [1/(xa)]. [This rule is true for any variable or expression, we just use x for convenience.]

Thus, for the first expression, the negative exponent causes it to "flip" into the bottom of a fraction.

x−4 =

1

x4

Things work similarly for the second expression as well. Because it has a negative exponent, the fraction flips, and then we apply the now-positive exponent to each part:

⎛⎝

4

7

⎞⎠

−2

=

⎛⎝

7

4

⎞⎠

2

=

72

42

=

49

16

We continue this process with the third expression, the only difference is that we want to deal with the negative exponents inside the parentheses first. The negative exponent on the u causes it to flip to the bottom of the fraction, while the negative exponent on the v causes it to flip to the top:

⎛⎝

u−4

v−6w2

⎞⎠

−3

=

⎛⎝

v6

u4w2

⎞⎠

−3

Once that's taken care of, we apply the −3 exponent on the outside of the parentheses to the entire fraction:

⎛⎝

v6

u4w2

⎞⎠

−3

=

⎛⎝

u4w2

v6

⎞⎠

3

=

u4·3w2·3

v6·3

=

u12w6

v18

[Alternatively, you could also use the rule of exponentiation on exponentiation causing exponents to multiply:

⎛⎝

u−4

v−6w2

⎞⎠

−3

=

u(−4) (−3)

v(−6)(−3)w2(−3)

=

u12

v18w−6

=

u12w6

v18

Either way is acceptable and gets you to the same answer.]

[1/(x4)] [49/16] [(u12w6)/(v18)]

Simplify each of the below expressions.

64[1/3] ( 257 )[1/14]

The important rule here is that fractional exponents are connected to roots: x[1/n] = n√{x}. [This rule is true for any variable or expression, we just use x for convenience.]

Thus, for the first expression, we get the appropriate root:

64[1/3] =

3

√

64

.

Remember, 3√{64} is the cube root of 64: that is, the number that, when raised to the third power, will give us 64. After thinking about it for awhile, we realize that 4·4 ·4 = 64, and so 3√{64} = 4.

For the next expression, we might be tempted to begin by figuring out what 257 is. Once we've used a calculator to find that out, we can attempt to take the 14 root of whatever giant number we get. The above would work, but there's a much easier way! Instead, begin by using one of our other rules: exponentiation on exponentiation causes the exponents to multiply.

( 257 )[1/14] = 257 ·[1/14] = 25[1/2]

At this point, it's much easier to apply the root rule:

25[1/2] =

√

25

= 5

[If you're wondering why it's written as √{25} and not 2√{25}, that's because the radical symbol (√{ }) is automatically assumed to be a square root (2√{ }) unless another number is put down to show that it is a different root.]

4 5

Approximate the below to the first two decimal places.

5π

It is impossible to write out the precise value of 5π with numbers because it will be an irrational number: one where the decimal expansion continues forever. However, we can get a good approximation by plugging in a decimal number for π, then using a calculator.

What number should we use for π? Remember, π is also irrational, so whatever number we use for it, it will also be an approximation. While we could use the approximation of 3.14, the less accurate our π approximation is, the less accurate our end result will be. Therefore, let's use (at least) the first six digits of π: 3.14159. If you want even more accuracy, use even more digits of π.

Now that we have a number that we can work with, we can plug it in to our calculator:

5π≈ 53.14159 = 156.9918748…

The problem asked for the first two decimal places, so we cut it off to 156.99. [Notice that 5π≠ 156.9918748…. Using a much longer string of digits for π, we can find that

5π = 156.9925….

Therefore, while 53.14159 = 156.9918748… is a good approximation, it is not perfect. In fact, no number can be perfect. Because we calculate the value using digits of π, we have to choose some specific number of digits. But since π goes on forever, we can only ever use an approximation of the true value of π, so our final result must also be an approximation. We can get very good approximations, but they can never be absolutely precise. The only way to represent what the number is absolutely precisely is by using what we started with: 5π.]

156.99 [If you got an answer that was close, but not quite the same as the above, you probably need to use more digits of π. Check out the steps for a more detailed discussion.]

Simplify the below expression.

(x2 ·x−10 ·x7 )5

(x80 ·x20 )−[1/5]

To do this, we will need to use a combination of the rules we've gone over previously in these questions. These rules can be applied in many different orders, but will all give the same result. The steps below will give a method that is fairly direct and quick, but they are not the only way to find the answer.

(x2 ·x−10 ·x7 )5

(x80 ·x20 )−[1/5]

=

(x2−10+7)5

(x80+20 )−[1/5]

=

(x−1)5

(x100 )−[1/5]

(x−1)5

(x100 )−[1/5]

=

x−1 ·5

x100 ·(− [1/5])

=

x−5

x−20

Since both the numerator and denominator each have negative exponents on them, they both "flip" sides:

x−5

x−20

=

x20

x5

At this point, we can just cancel out like usual: x20 means multiplying 20 x's, while x5 means multiplying 5 x's, so the 5 on the bottom will cancel out 5 of those on top:

x20

x5

= x15

x15

Simplify the below expression.

173 ·

2a⎛√

3−6a ·38a

36a ·176a

To do this, we will need to use a combination of the rules we've gone over previously in these questions. These rules can be applied in many different orders, but will all give the same result. The steps below will give a method that is fairly direct and quick, but they are not the only way to find the answer.

173 ·

2a⎛√

3−6a ·38a

36a ·176a

= 173 ·

2a⎛√

3−6a + 8a

36a ·176a

= 173 ·

2a⎛√

32a

36a ·176a

Notice that since 176a does not have the same base as 32a and 36a, it can't really interact with them directly: there is no way to combine them. For now, just set it aside from the fraction (although it must remain inside the radical).

To find (f °g ) (x), we compose the functions by plugging one into the other. Remember, it's almost always easier to know what to do by writing the above function composition in its equivalent form:

⎛⎝

f °g

⎞⎠

(x) = f

⎛⎝

g(x)

⎞⎠

Plug g(x) in to f:

f

⎛⎝

g(x)

⎞⎠

= f( x[3/4]) = 17 − ( x[3/4] )[8/9]

Simplify based on the rules of exponentiation:

17 − ( x[3/4] )[8/9] = 17 − x[3/4] ·[8/9] = 17 − x[2/3]

Thus, in simplest form, we have (f °g ) (x) = 17 − x[2/3].

To find the value (f °g ) (27), just plug x=27 in to what we just found:

⎛⎝

f °g

⎞⎠

(27) = 17 − (27)[2/3]

From there, just simplify based on the rules of exponentiation:

17 − (27)[2/3] = 17 −

⎛⎝

3

√

27

⎞⎠

2

= 17 − (3)2 = 17−9 = 8

(f °g ) (x) = 17 − x[2/3], (f °g ) (27) = 8

Let t be a number such that 64t = [9/4]. What is 8t?

Right now, we can't directly figure out what the value of t is. However, we can figure out how 8 connects to 64 in terms of exponentiation. [In a few lessons, once we learn about logarithms, we will be able to solve for t directly. But it would still be easier to do this problem the way shown below.]

We can connect 8 and 64 through exponentiation by the following:

√

64

= 8 ⇒ 64[1/2] = 8

Once we see that, we can replace 8 with 64[1/2].

Plugging in to our expression 8t, we have

8t = (64[1/2] )t = 64[1/2] ·t = 64t ·[1/2] = ( 64t )[1/2]

We can now replace 64t with what we were given in the problem, then simplify:

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Understanding Exponents

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.